3 Answers
3

Stirling's approximation gives us that
$$
n!=\sqrt{2\pi n}\ n^n\,e^{-n}(1+O(1/n))\tag{1}
$$
As it is simply the $j=k$ term in
$$
\sum_{j=0}^n\binom{n}{j}\left(\frac{k}{n}\right)^{j}\left(1-\frac{k}{n}\right)^{n-j}=1\tag{2}
$$
we can deduce that
$$
\binom{n}{k}\left(\frac{k}{n}\right)^{k}\left(1-\frac{k}{n}\right)^{n-k}\le1\tag{3}
$$
Therefore,
$$
\begin{align}
&\sum_{k=0}^n\frac{k^k(n-k)^{n-k}}{k!(n-k)!}n!n^{-n-1/2}\\
&=\sum_{k< m}\binom{n}{k}\left(\frac{k}{n}\right)^{k}\left(1-\frac{k}{n}\right)^{n-k}n^{-1/2}\\
&+\sum_{k=m}^{n-m}\binom{n}{k}\left(\frac{k}{n}\right)^{k}\left(1-\frac{k}{n}\right)^{n-k}n^{-1/2}\\
&+\sum_{k>n-m}\binom{n}{k}\left(\frac{k}{n}\right)^{k}\left(1-\frac{k}{n}\right)^{n-k}n^{-1/2}\\
&=O(mn^{-1/2})+\frac1{\sqrt{2\pi}}\sum_{k=m}^{n-m}\frac1{\sqrt{k}\sqrt{n-k}}(1+O(1/m))+O(mn^{-1/2})\\
&=O(mn^{-1/2})+\frac1{\sqrt{2\pi}}\sum_{k=0}^n\frac1{\sqrt{k/n}\sqrt{1-k/n}}\frac1n(1+O(1/m))\tag{4}
\end{align}
$$
We can add the tails to the last sum in $(4)$ since each of the $2m$ terms is $\le n^{-1/2}$ and absorb the difference into $O(mn^{-1/2})$.